Determination of Acetic Acid in Vinegar

The goal is to practice the use of standardizing a titrant and then using that solution to titrate an unknown in order determine the parts per trillion of acetic acid in vinegar and quantify it into grams of acetic acid per 100mL of vinegar.

Procedure:

1. Add 7.0 ml NaOH to a 1L plastic bottle and immediately fill the bottle with distilled water. Shake for several minutes until all is dissolved. Save any extra from this experiment.

2. Obtain ~4g of KHP’s purity factor. Weigh out three sample varying from 0.7 to 0.9g into 0.25L Erlenmeyer flasks. Dissolve with 50mL of distilled water. May heat to expedite mixing. Mix thoroughly. Add 2 drops of phenolphyhalen indictor solution and titrate. Very faint pink that stays for 20 seconds after thorough mixing is the end point.

3. Obtain vinegar sample. Pipet 25ml into a 0.25 into a volumetric flask, add water to mark and mix. Pipet 50mL of this into a 0.25L Erlenmeyer flask with 2 drop of phenolphthalein and 25mL of distilled water. Titrate other two samples. Rinse out buret.

4. Calculate and report concentration grams of acetic acid per 100mL of vinegar. Assume that all the acid present in vinegar to be acetic acid. Use standard deviation method from page 6 of Lab manual.

Chemicals to be used:

Chemical Name

Chemical Formula

Molecular Weight

State

Hazard Info

Sodium Hydroxide

NaOH

39.997 g/mol

Liquid

Alkalotic

Phenolphthalein

C20H14O4

318.32 g mol−1

Liquid

None

KHP

KC8H5O4

204.22 g mol−1

Solid

Slightly Acidic

Acetic Acid

C2H4O2

60.05 g mol−1

Liquid

Acidic

Principles:

In this lab the concept of acid and pKa’s are discussed and dealt with. pKa is the propensity of an acid to donate its hydrogen. The pKa is the negative logarithm of the Ka (which is the molarity of the liquid products over the molarity of the reactants). In the simple equation: HA ↔ H+ + A–. The moles of HA, if completely disassociated will equal the number of moles of the hydrogen and it will equal the moles of the conjugate base. So in order to titration the acetic acid the equation will be: CH3COOH + NaOH ↔ CH3COO– Na+ + H2O. So the moles of acetic acid will be titrated by the same number of moles of sodium hydroxide. This is under the assumption that the pKa of CH3COOH shows a high propensity to donate its hydrogen. The pKa of acetic acid is 4.76, which is on the threshold of being a strong acid (according to Harris page 181 in the margin).

Regardless, the main concentration of this experiment will be on the titration. A titration, according to Harris, is “a procedure in which one substance (titrant) is carefully added to another (analyte) until complete reaction has occurred. The quantity of titrant required for the complete reactions tells how much analyte is present” (Harris 565). In short, titration reveals how much analyte is present.

The buret is one of the main instruments for working a titration. It is cylindrical apparatus with the main purpose of adding small amount of a titrant to a solution at a time. The “stopcock” is this valve that allows the titrant to move drop wise into the solution. The buret is filled past the 0.00mL mark (in this case for our particular equipment) and then drained to the zero mark, this should be the initial starting point for the sake of simplicity. First the titrant needs to be standardized, so it becomes known what molarity the titrant is. This is done by taking a known standard, such as KHP, taking the mass, finding the moles, converting that via the stoichiometical ratio to the moles of the titrant all divided by the amount need to titrate the grams of KHP. In doing this, the molarity of titrant is found.

After this the titration of the unknown is done in quite the same manner as standardizing the titrant.

Titration is one of the many tools at access to the analytical chemist and is the basic starting foundation for this subject.

According to wwwchem.csustan.edu/consumer/vinegar/analysis.htm the federal requirement for acetic acid is 4g per 100mL of vinegar. The final value of this experiment was 7.65±0.05g/0.1L . So the percent error was 91.25%. This dramatic error would be better if the KHP standardization went better. It does not make sense that the first trial had a larger mass but a smaller volume than the second trial. If the molarity of NaOH had been 0.1M, the g per 0.1L vinegar sample would have been 4.04. Very close, very academically disgusting.

The weakest link in precision or accuracy was precision in the KHP standardization if is obvious because of the varied volume of NaOH added to titrate. The second trial values are definitely outliers. This skews the average g of KHP and mL of NaOH dramatically.